Problem: $\begin{aligned} &f(n)=2\left(\dfrac{1}{3}\right)^{n-1}+4 \\\\ &h(t)=\dfrac{t+4}{t-2} \end{aligned}$ $(h\circ f) (0)=$
Answer: Let's start by rewriting $(h\circ f) (0)$ as $h(f(0))$. When evaluating composite functions, we work our way inside out. To evaluate $h(f(0))$, let's first evaluate $f(0)$. Then we'll plug that result into $h$ to find our answer. Let's evaluate $f({0})$. $\begin{aligned}f(n)&=2\left(\dfrac{1}{3}\right)^{n-1}+4\\\\ f({0})&=2\left(\dfrac{1}{3}\right)^{({0})-1}+4~~~~~~~~~~\text{Plug in }n={0}\\\\ &=2\left(\dfrac{1}{3}\right)^{-1}+4\\\\ &=2(3)+4\\\\ &={10}\end{aligned}$ We now know that $h(f({0}))$ is the same as $h({10})$ because $f({0}) = {10}$. Let's evaluate $h({10})$. $\begin{aligned}h(t)&=\dfrac{t+4}{t-2}\\\\ h({{10}})&=\dfrac{({10})+4}{({10})-2}~~~~~~~~~~\text{Plug in }t={10}\\\\ &=\dfrac{14}{8}\\\\ &=\dfrac{7}{4}\\\\\end{aligned}$ The answer: $(h\circ f)(0) =\dfrac{7}{4}$